What is the integral of #int (cos2x)^2 dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Eddie Jul 20, 2016 #= 1/8 sin 4x + x/2 + C# Explanation: use the ID: #cos 2A = 2 cos^2 A - 1# or #cos^2 A = 1/2(cos 2A + 1)# the integral becomes #1/2 int cos4x + 1 \ dx# #= 1/2 (1/4 sin 4x + x) + C# #= 1/8 sin 4x + x/2 + C# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 13120 views around the world You can reuse this answer Creative Commons License